If omega is a complex cube root of unity,then | vedclass

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(A) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.First,simplify the powers of $\omega$:$

Vedclass · Accepted Answer

$1 / \sqrt{2}$

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(A) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.First,simplify the powers of $\omega$:$\omega^{10} = (\omega^3)^3 \cdot \omega = 1^3 \cdot \omega = \omega$$\omega^{23} = (\omega^3)^7 \cdot \omega^2 = 1^7 \cdot \omega^2 = \omega^2$Now substitute these into the expression:$\sin \left[ (\omega + \omega^2) \pi - \frac{\pi}{4} \right]$Since $1 + \omega + \omega^2 = 0$,it follows that $\omega + \omega^2 = -1$.Substituting this value:$\sin \left[ (-1) \pi - \frac{\pi}{4} \right] = \sin \left( -\pi - \frac{\pi}{4} \right)$$= \sin \left( -\frac{5\pi}{4} \right) = -\sin \left( \frac{5\pi}{4} \right)$$= -\sin \left( \pi + \frac{\pi}{4} \right) = -(-\sin \frac{\pi}{4}) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

If $\omega$ is a complex cube root of unity,then $\sin \left[\left(\omega^{10}+\omega^{23}\right) \pi-\frac{\pi}{4}\right]=$

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